Find k so thatis continous at x=5.

Question

Find k so that
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 25 over denominator straight x minus 5 end fraction comma space if space straight x not equal to 5 end cell row cell space space space space space space space straight k space space space space space space space comma space if space straight x equals 5 end cell end table close$
is continous at x=5.

Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 25 over denominator straight x minus 5 end fraction comma space if space straight x not equal to 5 end cell row cell space space space space space space space straight k space space space space space space space comma space if space straight x equals 5 end cell end table close space Lt with straight x rightwards arrow 5 below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 below fraction numerator straight x squared minus 25 over denominator straight x minus 5 end fraction equals Lt with straight x rightwards arrow 5 below fraction numerator left parenthesis straight x minus 5 right parenthesis left parenthesis straight x plus 5 right parenthesis over denominator straight x minus 5 end fraction equals Lt with straight x rightwards arrow 5 below left parenthesis straight x plus 5 right parenthesis equals 5 plus 5 equals 10 Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 5 therefore space space space space straight f left parenthesis 5 right parenthesis equals Lt with straight x rightwards arrow 5 below straight f left parenthesis straight x right parenthesis space space space space rightwards double arrow space straight k equals 10